[LeetCode] Regular Expression Matching

[LeetCode] Regular Expression Matching

 

Regular Expression Matching

 

Implement regular expression matching with support'.'And'*'.

'.' Matches any single character.'*' Matches zero or more of the preceding element.The matching should cover the entire input string (not partial).The function prototype should be:bool isMatch(const char *s, const char *p)Some examples:isMatch("aa","a") → falseisMatch("aa","aa") → trueisMatch("aaa","aa") → falseisMatch("aa", "a*") → trueisMatch("aa", ".*") → trueisMatch("ab", ".*") → trueisMatch("aab", "c*a*b") → true

Solution:

The question is to implement regular expression matching. Supported. And *

There are three cases. Set the subscript of the current string and mode to I, j, respectively.

1. If the current mode is matched, p [j] = '\ 0'. If the string ends, true is returned; otherwise, false is returned.

2. If the next character in the mode is not *, p [j + 1]! = '*'. We will discuss the situation here.

(1) If s [I] = p [j], recursive verification of I + 1, j + 1

(2) If p [I] = '.' And s [I]! = '\ 0', recursive verification I + 1, j + 1

(3) Otherwise, false is returned.

3. If the next character of the mode is *, that is, p [j + 1] = '*', Recursive Backtracking is performed continuously, j + 2 (k increases from 0 to len (s)-I, j + 2 means crossing the current character and *).

The Code is as follows:

 

class Solution {public:    bool isMatch(string s, string p) {        return matchHelper(s, p, 0, 0);    }        bool matchHelper(string& s, string& p, int i, int j){        if(p[j]=='\0'){            return s[i]=='\0';        }                if( p[j + 1] != '*'){            return ((s[i] == p[j]) || (p[j] == '.' && s[i]!='\0')) && matchHelper(s, p, i + 1, j + 1);        }                while((s[i] == p[j]) || (p[j] == '.' && s[i]!='\0')){            if(matchHelper(s, p, i, j+2)) return true;            i++;        }        return matchHelper(s, p, i, j+2);    }};